Question

An LR-series circuit has a variable inductor with the inductance defi­ned by

$\mathit{L}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}1-\frac{1}{10}t,0\le t\le 10\\ 0,t>10\end{array}$

Find the current i(t) if the resistance is 0.2 ohm, the impressed voltage is E(t) 5 4, and i(0) 5 0. Graph i(t).

Short answer

$i\left(t\right)=\left\{\begin{array}{l}20-\frac{1}{5}{\left(10-t\right)}^2,0\le t\le 10\\ 20,t>10\end{array}\right.$

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$\mathit{L}\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{R}\mathit{i}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

We have an impressed voltage as $E=4$volts in a circuit in series with a variable inductance

$i\left(t\right)=\left\{\begin{array}{l}1-\frac{1}{10}t,0\le t\le 10\\ 0,t>10\end{array}\right.$

and resistance $R=0.2\text{ohms},$and then we have the differential equation for current as

$L\frac{di}{dt}+Ri=E\left(t\right)$-------- (1)

and we have to obtain the current I(t) of this circuit if we have the initial current as $I\left(0\right)=0$as the following :

For the interval $0⩽t⩽10$: since we have $L=1-\frac{1}{10}t$henry, $R=0.2$ohms and $E=4$volts, then we can have the differential equation shown in (1) as

$\begin{array}{rcl}\left(1-\frac{1}{10}t\right)\frac{di}{dt}+0.2i& =& 4\\ \left(\frac{10-t}{10}\right)\frac{di}{dt}+0.2i& =& 4\times \frac{10}{10-t}\\ \frac{di}{dt}+\frac{2}{10-t}i& =& \frac{40}{10-t}\\ \frac{di}{dt}& =& \frac{40}{10-t}-\frac{2}{10-t}i\end{array}$

$\frac{di}{dt}=\frac{40-2i}{10-t}$-------- (2)

This equation is a first order linear and separable differential equation, then we can separate variables and then do integration as

Separate the variables

$\begin{array}{rcl}\frac{di}{40-2i}& =& \frac{dt}{10-t}\\ -\frac{1}{2}\int \frac{-2}{40-2i}di& =& -\int \frac{-1}{10-t}dt\\ -\frac{2}{-}\ln (40-2i)& =& -\ln (10-t)+\ln c\\ \ln (40-2i)& =& 2\ln (10-t)+\ln k\\ \ln (40-2i)& =& \ln \left(k{(10-t)}^2\right)\end{array}$

Then we have

$\begin{array}{rcl}{e}^{\ln (40-2i)}& =& e^{\ln \left(k{(10-t)}^2\right)}\\ 40-2i& =& k(10-t{)}^2\\ 2i& =& 40-k(10-t{)}^2\end{array}$

Then we obtain

$\begin{array}{rcl}i\left(t\right)& =& \frac{40-k{(10-t)}^2}{2}\\ & =& 20-\frac{1}{2}k(10-t{)}^2\end{array}$ -------- (3)

After that, to find the value of constant we have to apply this point of condition $(i,t)=(0\text{amp},0\text{seconds)}$into equation (3) as

$\begin{array}{rcl}0& =& 20-\frac{1}{2}k(10-0{)}^2\\ 50k& =& 20\end{array}$

Then we have

$k=\frac{2}{5}$

Substitution

After that, substitute with the value of into equation (3), then we have

$i\left(t\right)=20-\frac{1}{5}(10-t{)}^2$-------- (4)

is the current of LR series circuit for the interval $0⩽t⩽10.$

Now for the interval $t>20$: since we have $L=0,R=0.2\text{ohms}$and $E=4$volts, then we can have equation (1) as

$\begin{array}{rcl}\left(0\right)\frac{di}{dt}+0.2i& =& 4\\ \frac{1}{5}i& =& 4\end{array}$

Then we have

$\begin{array}{rcl}i& =& 4\times 5\\ & =& 20\text{amp}\end{array}$ -------- (5)

is the current of series circuit for the interval$t>10.$

Finally, from equations (4) and (5), we can obtain the current of series circuit as

$i\left(t\right)=\left\{\begin{array}{l}20-\frac{1}{5}{\left(10-t\right)}^2,0\le t\le 10\\ 20,t>10\end{array}\right.$